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Question

A force F=(5^i+3^j+2^k) N is applied over a particle which displaces it to a point having position vectorr=(2^i^j) m from origin. The work done on the particle (in J) is

A
+10
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B
+7
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C
7
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D
+13
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Solution

The correct option is B +7
Given constant force F=(5^i+3^j+2^k) N
Initial position r1=(0^i+0^j+0^k) m, as starting from origin.
Final position r2=(2^i^j+0^k) m

To find the displacement s:
s=r2r1= change in position
s=(2^i^j+0^k)(0^i+0^j+0^k)
s=(2^i^j) m

Hence the work done , W=F.S
W=(5^i+3^j+2^k).(2^i^j)=103=7 J

Hence option B is the correct answer.

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